Question #eaa82

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Question #eaa82

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Answer 1 · 2016-12-02T02:31:24

Approx. $4 \cdot g$ $4*g$

Explanation:

The percentage mass of magnesium in magnesium carbonate, $M g C O_{3}$ $MgCO_3$ :

$= \frac{Molar mass of metal}{Molar mass of magnesium carbonate} = \frac{24.305 \cdot g \cdot m o l^{- 1}}{84.32 \cdot g \cdot m o l^{- 1}} \times 100 % = 28.8 %$ $="Molar mass of metal"/"Molar mass of magnesium carbonate"=(24.305*g*mol^-1)/(84.32*g*mol^-1)xx100%=28.8%$

And thus in a $13.9 \cdot g$ $13.9*g$ sample, there are $28.8 % \times 13.9 \cdot g ≅ 4 \cdot g$ $28.8%xx13.9*g~=4*g$

Answer 2 · 2016-12-02T03:56:04

The Answer is approximately
$≅ 4 g$ $~= 4 g$

Explanation:

There is another way to consider this question Atomic wt of all atoms in the compound are as follows:
$C = 12, M g = 24, O = 16$ $C=12, Mg=24, O=16$
Wt of compound = $C M g O 3 = 12 + 24 + 16 \cdot (3) = 84$ $CMgO3 = 12 + 24 + 16*(3) = 84$
It means 84 gm CMgO3 gives 24 gm Magnesium
Therefore 13.9 g CMgO3 gives $\frac{24}{84} \cdot 13.9 = 3.97 ≅ 4 g m$ $24/84 *13.9 = 3.97 ~=4 gm$

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